Question

A 39.3 g glass thermometer reads 22.0oC before it

isplaced in 136 mL of water. When the water and thermometer come
toequilibrium, the thermometer reads 43.6oC. What was
theoriginal temperature of the water in oC? Do not
enterunit.

Answer 1

44.85C

Step-by-step explanation:

Let the specific heat of glass thermometer be 0.84 J/g°C

Let the specific heat of water be 4.186 j/g °C

Let the water density be 1kg/L

136 mL of water = 0.136L of water = 0.136 kg of water = 136 g of water

Since the change of temperature on the glass thermometer is 43.6 - 22 = 21.6 C. We can then calculate the heat energy absorbed to it:

`E = m_gc_g \Delta T = 39.3 * 0.84 * 21.6 = 713.06 J`

Assume no energy is lost to outside, by the law of energy conservation, this heat energy would come from water

`E = m_wc_w(T - T_w) = 713.06`

`136*4.186(T - 43.6) = 713.06`

`T - 43.6 = (713.06)/(136*4.186) = 1.25`

`T = 1.25 + 43.6 = 44.85C`